The construction of a Bernstein set \(B\subseteq {{R}}\)  provided in [1]} works equally well for \(B\subseteq {{N}}^{{N}}\) 
and is by transfinite recursion.
This construction necessarily requires the axiom of choice.
Indeed, if we assume \(\mathsf {BP}\) , i.e., that every subset \(B\subseteq {{N}}^{{N}}\)  has the Baire property,
then no Bernstein set can exist [1]}.
From the point of view of computability theory Bernstein sets play a similar rôle as immune sets.
Actually, together with Proposition REF  and Fact REF 
we directly obtain the following conclusion.
